I am given that X is a standard normal distribution.
Why is $Cov(X, -X) = -1$?
I know that $Cov(X,X) = Var(X)$ and that the $Var(X) = 1$.
Is the $Var(-X) = -1$?
This follows from $Cov(X,-X)=-Cov(X,X)$. In fact, more generally, $Cov(aX,bY)=abCov(X,Y)$.
Also, the above property can be used to show $Var(-X)=(-1)^2Var(X)=Var(X)$ so $Var(-X)\neq -1$ (in fact, the variance is always nonnegative).